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 August 21st, 2013, 09:41 PM #1 Member   Joined: Aug 2013 Posts: 69 Thanks: 0 Algebra Check UP! WHY am I OFF by sqrt. 2 1/?((1-((x-1)/(x+1))^2 ) )*((1+x)-(1-x))/(1+x)^2 I end up getting = 2/?2x(1+x) ...the answer says it should 1/?x(1+x) ...Yes I did use the variable 2. And the answer was the same as the book... I've tried, but am blocked. Zen.
 August 22nd, 2013, 12:33 AM #2 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: Algebra Check UP! WHY am I OFF by sqrt. 2 $\frac {1}{\sqrt {1-\frac {(x-1)^2}{(x+1)^2} }}*\frac {(1+x)-(1-x)}{(1+x)^2}$ $\frac {2x}{(1+x)^2 \, \sqrt {1-\frac {(x-1)^2}{(x+1)^2} }}$ $\frac {2x}{(1+x) \, \sqrt {(1+x)^2-(x-1)^2 }}$ $\frac {2x}{(1+x) \, \sqrt {2x*2}}$ $\frac {x}{(1+x) \, \sqrt {x}}$
 August 22nd, 2013, 10:34 PM #3 Newbie   Joined: May 2013 Posts: 16 Thanks: 0 Re: Algebra Check UP! WHY am I OFF by sqrt. 2 Solution: 1/(?(1-((x-1)²/(x+1)²))*((1+x)-(1-x))/(1+x)² =2x/((1+x)²?((x+1)²-(x-1)²/(x+1)²) =2x/((x+1)?((x+1)²-(x-1)²)) =2x/(x+1)?(4x) =x/(x+1)?x =?x/(x+1) this is the right answer.
 August 24th, 2013, 01:16 PM #4 Member   Joined: Aug 2013 Posts: 69 Thanks: 0 Re: Algebra Check UP! WHY am I OFF by sqrt. 2 Okay, I should go to sleep more , I have to brush up on my easy arithmetic, x^2 + 2x + 1 (not x^2 + x + 1) Thanks.
 August 24th, 2013, 01:20 PM #5 Member   Joined: Aug 2013 Posts: 59 Thanks: 0 Don't be lazy to learn the benefit of LaTeX. Thanks to gelatine1, we could move on to solve such problem. $\frac{x}{\sqrt{x}}\,=\,\frac{x^{\frac{2}{2}}}{x^{\ frac{1}{2}}}\,=\,x^{\frac{2}{2}\,-\,\frac{1}{2}}\,=\,x^{\frac{1}{2}}\,=\,\sqrt{x}.$

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